Problem: The lifespans of gorillas in a particular zoo are normally distributed. The average gorilla lives $18.8$ years; the standard deviation is $3.8$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a gorilla living longer than $11.2$ years.
Solution: $18.8$ $15$ $22.6$ $11.2$ $26.4$ $7.4$ $30.2$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $18.8$ years. We know the standard deviation is $3.8$ years, so one standard deviation below the mean is $15$ years and one standard deviation above the mean is $22.6$ years. Two standard deviations below the mean is $11.2$ years and two standard deviations above the mean is $26.4$ years. Three standard deviations below the mean is $7.4$ years and three standard deviations above the mean is $30.2$ years. We are interested in the probability of a gorilla living longer than $11.2$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the gorillas will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the gorillas will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $11.2$ years and the other half $({2.5\%})$ will live longer than $26.4$ years. The probability of a particular gorilla living longer than $11.2$ years is ${95\%} + {2.5\%}$, or $97.5\%$.